First slide

Beats

Question

Two wires are fixed in a sonometer. Their tensions are in the ratio 8 : 1. The lengths are in the ratio 36 : 35. The diameters are in the ratio 4 : 1. Densities of the materials are in the ratio 1 : 2. If the lower frequency in the setting is 360 Hz. the beat frequency when the two wires are sounded together is

Difficult

A

5
B

8
C

6
D

10

Solution

Frequency in a stretched string is given by $n = \frac{1}{2 l} \sqrt{\frac{T}{{πr}^{2} ρ}} = \frac{1}{l} \sqrt{\frac{T}{{πd}^{2} ρ}}$ (d = Diameter of string)
$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{l_{2}}{l_{1}} \sqrt{\frac{T_{1}}{T_{2}} \times {(\frac{d_{2}}{d_{1}})}^{2} \times (\frac{ρ_{2}}{ρ_{1}})}$
$= \frac{35}{36} \sqrt{\frac{8}{1} \times {(\frac{1}{4})}^{2} \times \frac{2}{1}} = \frac{35}{36} \Rightarrow n_{2} = \frac{36}{35} \times 360 = 370$
Hence beat frequency = $n_{2} - n_{1} = 10$

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